ap chapter 9. why is it easier to open the door from the outer edge instead of closer to the hinges?

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AP Chapter 9

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Page 1: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

AP Chapter 9

Page 2: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• Why is it easier to open the door from the outer edge instead of closer to the hinges?

Page 3: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• Your force is applied farther from the pivot point resulting in a larger lever arm.

• The lever arm combined with the force creates a new phenomena called torque

• τ = Fl τ is called tau and units are Nm

Page 4: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• What is the net torque on a 4 m teeter totter if you, 58 kg, and your friend, 45 kg, are on opposite ends with the fulcrum (pivot) in the center?

Page 5: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• τ =(58kg)(9.8m/s2)(2m) –(45kg)(9.8m/s2)(2m)

• τ = 255 Nm

0 m2m 2m

58 kg 45 kg

Page 6: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• In equilibrium

• ΣFx = 0 Σ Fy = 0 Σ τ = 0

Page 7: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• Steps to solve• 1) Free-body diagram• 2) Choose convenient axes• 3) Apply the force equations• 4) Choose convenient pivot point• 5) Apply torque equation• 6) Solve

Page 8: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• If you, 55 kg, climb a 6.0 m, 25 kg ladder and stand on the rung that is 5.6 m up and the ladder is 40o from the ground. How much force does the wall apply to the ladder assuming no friction with the wall? How about the ground?

Page 9: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

Fw

Fyou

Flad

Fgy

Fgx

6.0m

5.6m

3.0m

40o

Page 10: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• ΣFx = Fgx – Fw = 0 Fgx = Fw

• ΣFy = Fgy – Flad – Fyou = 0• Fgy = Flad + Fyou = • Fgy = (25kg)(9.8m/s2) + (55kg)(9.8m/s2)• Fgy = 784 N

Page 11: AP Chapter 9. Why is it easier to open the door from the outer edge instead of closer to the hinges?

• Στ = Fw(6.0m)sin40o – Fyou(5.6m)sin50o – Flad(3.0m)sin50o = 0

• Fw(6.0m)sin40o= (55kg)(9.8m/s2)(5.6m) sin50o + (25kg)(9.8m/s2)(3.0m)

sin50o

• Fw = 750 N

• Fgx = Fw = 750 N